Rng（逆元）

Rng

Problem Description
Avin is studying how to synthesize data. Given an integer n, he constructs an interval using the following method: he first generates a integer r between 1 and n (both inclusive) uniform-randomly, and then generates another integer l between 1 and r (both inclusive) uniform-randomly. The interval [l, r] is then constructed. Avin has constructed two intervals using the method above. He asks you what the probability that two intervals intersect is. You should print p* q(−1)(MOD 1, 000, 000, 007), while pq denoting the probability.

Input
Just one line contains the number n (1 ≤ n ≤ 1, 000, 000).

Output
Print the answer.

Sample Input

1
2

Sample Output

1
750000006

这题首先找到规律之后用逆元求解，规律为（n+1）/2n

求相交的概率就是求1-不相交的概率，因为两个区域一定是一个在左边一个在右边，从右边往左边枚举，分别有n-1，n-2，n-3,…,3,2,1,所以用求和公式得到n（n-1）/2 再除以一个他们的总的可能性n²,得到上面这个公式。

Code：

#include<iostream>
#include<vector>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;
const int p=1e9+7;
ll quickpow(ll a,ll b)
{

    ll ret=1;
    while(b!=0)
    {
        if(b&1)ret=ret*a%p;
        b>>=1;
        a=a*a%p;
    }
    return ret%p;
}
int main()
{
    int n;
    while(cin>>n)
    {
        ll ans=(n+1)*quickpow(2*n,p-2)%p;
        printf("%lld\n",ans);
    }
    return 0;
}